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Variation of Parameters

and we saw that while it reduced things down to just an
algebra problem, the algebra could become quite messy. On top of that undetermined coefficients will
only work for a fairly small class of functions.

The method of Variation of Parameters is a much more general
method that can be used in many more cases.
However, there are two disadvantages to the method. First, the complementary solution is
absolutely required to do the problem.
This is in contrast to the method of undetermined coefficients where it
was advisable to have the complementary solution on hand, but was not
required. Second, as we will see, in
order to complete the method we will be doing a couple of integrals and there
is no guarantee that we will be able to do the integrals. So, while it will always be possible to write
down a formula to get the particular solution, we may not be able to actually
find it if the integrals are too difficult or if we are unable to find the
complementary solution.

We’re going to derive the formula for variation of
parameters. We’ll start off by acknowledging
that the complementary solution to (1) is

Remember as well that this is the general solution to the
homogeneous differential equation.

Also recall that in order to write down the complementary
solution we know that y1(t)
and y2(t) are a
fundamental set of solutions.

What we’re going to do is see if we can find a pair of
functions, u1(t) and u2(t) so that

will be a solution to (1). We have two unknowns here and so we’ll need
two equations eventually. One equation
is easy. Our proposed solution must
satisfy the differential equation, so we’ll get the first equation by plugging
our proposed solution into (1). The second equation can come from a variety
of places. We are going to get our
second equation simply by making an assumption that will make our work easier. We’ll say more about this shortly.

So, let’s start. If
we’re going to plug our proposed solution into the differential equation we’re
going to need some derivatives so let’s get those. The first derivative is

Here’s the assumption.
Simply to make the first derivative easier to deal with we are going to
assume that whatever u1(t)
and u2(t) are they will
satisfy the following.

(3)

Now, there is no reason ahead of time to believe that this
can be done. However, we will see that
this will work out. We simply make this
assumption on the hope that it won’t cause problems down the road and to make
the first derivative easier so don’t get excited about it.

We’ve almost got the two equations that we need. Before proceeding we’re going to go back and
make a further assumption. The last
equation, (4),
is actually the one that we want, however, in order to make things simpler for
us we are going to assume that the function p(t) = 1.

In other words, we are going to go back and start working
with the differential equation,

If the coefficient of the second derivative isn’t one divide
it out so that it becomes a one. The
formula that we’re going to be getting will assume this! Upon doing this the two equations that we
want so solve for the unknown functions are

Note that in this system we know the two solutions and so
the only two unknowns here are and . Solving this system is actually quite
simple. First, solve (5)
for and plug this into (6) and
do some simplification.

Recall that y1(t)
and y2(t) are a
fundamental set of solutions and so we know that the Wronskian won’t be zero!

Finally, all that we need to do is integrate (8)
and (9)
in order to determine what u1(t)
and u2(t) are. Doing this gives,

So, provided we can do these integrals, a particular
solution to the differential equation is

So, let’s summarize up what we’ve determined here.

Variation of
Parameters

Consider the differential equation,

Assume that y1(t)
and y2(t) are a
fundamental set of solutions for

Then a particular solution to the nonhomogeneous
differential equation is,

Depending on the person and the problem, some will find the
formula easier to memorize and use, while others will find the process used to
get the formula easier. The examples in
this section will be done using the formula.

Before proceeding with a couple of examples let’s first
address the issues involving the constants of integration that will arise out
of the integrals. Putting in the constants
of integration will give the following.

The final quantity in the parenthesis is nothing more than
the complementary solution with c1
= -c and c2 = k and
we know that if we plug this into the differential equation it will simplify
out to zero since it is the solution to the homogeneous differential
equation. In other words, these terms
add nothing to the particular solution and so we will go ahead and assume that c = 0 and k = 0 in all the examples.

One final note before we proceed with examples. Do not worry about which of your two
solutions in the complementary solution is
y1(t) and which one
is y2(t). It doesn’t matter. You will get the same answer no matter which
one you choose to be y1(t)
and which one you choose to be y2(t).

Let’s work a couple of examples now.

Example 1 Find
a general solution to the following differential equation.

Solution

First, since the formula for variation of parameters
requires a coefficient of a one in front of the second derivative let’s take
care of that before we forget. The
differential equation that we’ll actually be solving is

We’ll leave it to you to verify that the complementary
solution for this differential equation is

So, we have

The Wronskian of these two functions is

The particular solution is then,

The general solution is,

Example 2 Find
a general solution to the following differential equation.

Solution

We first need the complementary solution for this
differential equation. We’ll leave it
to you to verify that the complementary solution is,

So, we have

The Wronskian of these two functions is

The particular solution is then,

The general solution is,

This method can also be used on non-constant coefficient
differential equations, provided we know a fundamental set of solutions for the
associated homogeneous differential equation.

Example 3 Find
the general solution to

given that

form a fundamental set of solutions for the homogeneous
differential equation.

Solution

As with the first example, we first need to divide out by
a t.

The Wronskian for the fundamental set of solutions is

The particular solution is.

The general solution for this differential equation is.

We need to address one more topic about the solution to the
previous example. The solution can be
simplified down somewhat if we do the following.

Now, since is an unknown constant subtracting 2 from it
won’t change that fact. So we can just
write the as and be done with it. Here is a simplified version of the solution
for this example.